How far up the slope will the block get and will it come back down?

[nightshade123]

Homework Statement

A block is shoved up a 22 deg slope with an Initial speed of 1.4 m/s the coefficient of Kinetic friction is .7...

a) how far up the slope will hte block get?

b) will the block come back down?note the bold is just a sign change, a diffrent possibility

2. Homework Equations and The attempt at a solution

x: f*g*sintheta+Friction of the surface = MAsubx

y: n- Force of g * costheta = 0

solve for Asubx and N

put n into the Ax eqn

Asubx = (m*g * sintheta - myuk * m * g * costheta) / m

m cancels factor out g

and use

v^2 = v0^2 + 2a(x-x0)v^2 = v0^2 + (2 ( g (sin theta)- myuk *costheta)))*(x-x0)

v = 0 m/s
v0= 1.4 m/s
g = 9.8 m/s^2
theta = 22 degrees
myuk = .7
x0 = 0 m
x = ? m

i get .364 m for part a and i said no it will not return back down due to a small incline and small v0

i THINK I am right but I am not 100% confident

note: i have a free body diagram and such ill make on in PAINT if u want me to, let me know, thanks for your time!

 

[Dick]

It looks fine to me. It won't come back down because the force of kinetic friction is greater than the gravitational force. What's the .1412m supposed to be?

 

[nightshade123]

i deleted the bold parts i just rememberd that it is opposite so its negitive,

 

[Dick]

The force of friction and the gravitational force are acting in the same direction when the block is going up the ramp. If you are using g=+9.8 (and it looks like you are) then the plus sign was the right one. Both terms should have the same sign.

 

[nightshade123]

so it was the .124 or so?

 

[nightshade123]

v^2 = v0^2 + (2 ( g (sin theta)+ myuk *costheta)))*(x-x0)

and yea i have g as positive and i noticed that after i calculated it
i guess just filling the in ? will answer my question

i believe its sintheta - myuk

 

[Dick]

2*(g*sin(theta)+mu*g*cos(theta)), right?

 

[nightshade123]

the part we are of interest right now we are trying to make sure is right is...

Asubx = Fsubg*sintheta PLUS OR MINUS Myu * NormalForceWHERE the normal force IS Fsubg * costheta

Fsubg =m * g

this was pretty close to my freebody diagram not exact... i think my angle might be in the wrong spot let me know

can u make a FBD and compare to mine,... i actauly think mine is wrong now that i stare at it for a min lol

 

[Dick]

Add them. The frictional force points in the opposite direction to the direction of motion. The gravitational force is always down the ramp. When the block is going up, they point in the same direction. When they are going down then they would point in opposite directions, but remember that if the magnitude of the static frictional force (which is greater than the kinetic frictional force) is greater than the gravitational force, then it won't slide down at all.

 

[nightshade123]

no static friction force was given.. but i made an assumption there is not way it is going to return down, is .7 high for a kinetic force? what is the equivalent to, we just started work with friction this past week, this was my first of a few real problems

Fsubg*sintheta PLUS Myu * NormalForcewith that said

v = 0 m/s
v0= 1.4 m/s
g = Minus 9.8 m/s^2
theta = 22 degrees
myuk = .7
x0 = 0 m
x = ? m

this should be obviously negitive but i want to see what you haev to say about it

Evaluate
v^2 = v0^2 + (2 ( -g (sin theta + myuk *costheta)))*(x-x0)

CORRECTION MY CALC WAS IN RADS LOL

the answer is .097691 m

 

[nightshade123]

alright this makes perfect sense if g is positive my answer is neg, so yea its soposed to be neg, if my math is right then that is the correct answer

 

[Dick]

v=0, and v0=1.4m/sec. Yes, the 2*a*d term should be negative if you write the equation in that form. But I don't know how you are getting .141m. Just use your intuition about what direction things should point to adjust the signs. What do you get for g*sin(22)+0.7*g*cos(22)?

 

[nightshade123]

-9.8 * (sin(22) + cos(22) * .7)
-9.8 * (1.02364)
-10.0316at that point in the eqn it is

0 = 1.96 - 2 ( 10.0316 ) *x

-1.96 = -20.0633 * x

x = .097691 m igonore the opening answer that is the real answer as we have moved through the problem i got in bold above

 

[Dick]

I agree with that.

 

[nightshade123]

thanks i appreciate it, i had a feeling about that + or minus sign, in my conlcusion i talked about how that could have thrown my answer off... .3m did sound liek a big distance for a 1.4 m/s velocity and the small i was making x come out i knew i was on the right path

take care have a good night =D